$x(e^x - e^{-x})-e^x = 0$ has two reals solutions but am I supposed to use the intermediate value theorem?
Solution 1:
Your problem is equivalent than showing that $$f(x):=x(1-e^{-2x})=1,$$ has several solutions. It's rather clear that $f(0)=0$ and $f(x)\to +\infty$ whenever $x\to \pm \infty $. So, oblviously, $f(x)=1$ has at least two solutions. It's also quite clear that $f(x)=1$ has exactly two solutions since $f'(x)<0$ whenever $x<0$ and $f(x)>0$ whenever $x>0$.